Relative atomic mass of NaCl=23+35.5=58.5
2000kg / 58.5 = mass per relative atomic mass = 34.188
34.188 * atomic mass of Na, 34.188 * 23 = 786.325... = 790 kg (2s.f.)
Normal galixies support millions of stars that support billions of planets. quesars are more similar to black holes. quesars also have gravitational pull
![[\text{Y}] \approx0.337\;\text{mol}\cdot\text{dm}^{-3}](https://tex.z-dn.net/?f=%5B%5Ctext%7BY%7D%5D%20%5Capprox0.337%5C%3B%5Ctext%7Bmol%7D%5Ccdot%5Ctext%7Bdm%7D%5E%7B-3%7D)
at equilibrium.
<h3>Explanation</h3>
Concentration for each of the species:
There was no Y to start with; its concentration could only have increased. Let the change in ![[\text{Y}]](https://tex.z-dn.net/?f=%5B%5Ctext%7BY%7D%5D)
be 
.
Make a 
table.
Two moles of X will be produced and two moles of Z consumed for every one mole of Y produced. As a result, the <em>change</em> in ![[\text{X}]](https://tex.z-dn.net/?f=%5B%5Ctext%7BX%7D%5D)
will be 
and the <em>change</em> in ![[\text{Z}]](https://tex.z-dn.net/?f=%5B%5Ctext%7BZ%7D%5D)
will be 
.
.
Add the value in the C row to the I row:
.
What's the equation of 
for this reaction? Raise the concentration of each species to its coefficient. Products go to the numerator and reactants are on the denominator.
![K_c = \dfrac{[\text{Z}]^{2}}{[\text{X}]^{2} \cdot[\text{Y}]}](https://tex.z-dn.net/?f=K_c%20%3D%20%5Cdfrac%7B%5B%5Ctext%7BZ%7D%5D%5E%7B2%7D%7D%7B%5B%5Ctext%7BX%7D%5D%5E%7B2%7D%20%5Ccdot%5B%5Ctext%7BY%7D%5D%7D)
.
. As a result,
![\dfrac{[\text{Z}]^{2}}{[\text{X}]^{2} \cdot[\text{Y}]} = \dfrac{(3-2x)^{2}}{(2+2x)^{2} \cdot x} = K_c = 2.25](https://tex.z-dn.net/?f=%5Cdfrac%7B%5B%5Ctext%7BZ%7D%5D%5E%7B2%7D%7D%7B%5B%5Ctext%7BX%7D%5D%5E%7B2%7D%20%5Ccdot%5B%5Ctext%7BY%7D%5D%7D%20%3D%20%5Cdfrac%7B%283-2x%29%5E%7B2%7D%7D%7B%282%2B2x%29%5E%7B2%7D%20%5Ccdot%20x%7D%20%3D%20K_c%20%3D%202.25)
.
.
The degree of this polynomial is three. Plot the equation 
on a graph and look for any zeros. There's only one zero at 
. All three concentrations end up greater than zero.
Hence the equilibrium concentration of Y: 
.
Answer:
165.5 g of CO2
Explanation:
We must first put down the balanced reaction equation:
C4H10(g) + 13/2 O2(g) ------> 4CO2(g) + 5H2O(g)
From the reaction equation, one mole of butane occupies 22.4 L hence we can establish the stoichiometry of the reaction thus:
22.4 L of butane created 174 g of CO2
Therefore 21.3 L of butane will create 21.3 × 174/22.4 = 165.5 g of CO2
;
;
.